Space of vector measures equipped with the total variation norm is complete

Let $$\Omega$$ be a set $$\mathcal A\subseteq2^\Omega$$ with $$\emptyset\in\mathcal A$$, $$E$$ be a $$\mathbb R$$-Banach space and $$\mu:\mathcal A\to E$$ be additive. Now, for $$A\subseteq\Omega$$, let $$|\mu|(A):=\sup\left\{\sum_{i=1}^k\left\|\mu(A_i)\right\|_E\right\},$$ where the supremum is taken over all $$k\in\mathbb N$$ and mutually disjoint $$A_1,\ldots,A_k\in\mathcal A$$ with $$\bigcup_{i=1}^kA_i\subseteq A$$.

It's easy to see that $$\mu\mapsto|\mu|(\Omega)\tag1$$ is a norm on the vector space of those $$\mu$$ for which $$|\mu|(\Omega)<\infty$$. Are we able to show that this norm is complete?

Assuming $$(\mu_n)_{n\in\mathbb N}$$ is a Cauchy sequence wrt $$(1)$$ of such $$\mu$$. For $$\varepsilon>0$$, there is a $$N\in\mathbb N$$ with $$|\mu_m-\mu_n|(\Omega)<\varepsilon\;\;\;\text{for all }m,n\ge N\tag2.$$ We should clearly have $$\left||\mu_m|(A)-|\mu_n|(A)\right|\le|\mu_m-\mu_n|(A)\le|\mu_m-\mu_n|(\Omega)\tag3.$$ So, $$(|\mu_n|(A))_{n\in\mathbb N}$$ is Cauchy.

If $$E=\mathbb R$$ this might help (a signed measure can be decomposed into a negative and a positive part), but I don't see what we need to do in general.

BTW: It would be great if someone knows a textbook reference where it is shown that the space of $$E$$-valued vector measures equipped with the total variation norm is complete.

• Do the inequalities in (3) require a decomposition theorem? It looks like they should be straightforward. – Umberto P. Apr 22 at 17:24

This isn't too different from the proofs that any of the basic function spaces are complete. Indeed suppose that $$(\mu_n)_{n \geq 1}$$ is Cauchy for $$| \cdot |(\Omega)$$. Let $$A_i \in \mathcal{A}$$. Then $$\| \mu_n(A_i) - \mu_m(A_i) \| \leq | \mu_n - \mu_m |(\Omega)$$ so $$\mu_n(A_i)$$ is a Cauchy sequence in the Banach space $$E$$ and hence converges to some element of $$E$$, which we will call $$\mu(A_i)$$. So we now have a function $$\mu: \mathcal{A} \to E$$ which is the pointwise limit of the sequence $$\mu_n$$. It is hence immediate that $$\mu$$ is finitely additive.
It remains to see that $$|\mu|(\Omega) < \infty$$ and that $$|\mu_n - \mu|(\Omega) \to 0$$. Let's prove them in that order. For the first, take arbitrary disjoint $$A_1, \dots A_k \in \mathcal{A}$$. Since $$\mu_n \to \mu$$ pointwise, for $$n$$ large enough $$\sum_i \| \mu_n(A_i) - \mu(A_i) \| \leq 1$$. Then, we can estimate, $$\sum_{i=1}^k \|\mu(A_i)\| \leq 1 + \sum_i \|\mu_n(A_i)\| \leq 1 + \sup_n |\mu_n|(\Omega) < \infty$$ since Cauchy sequences are bounded. So taking the $$\sup$$ on the left hand side gives us that $$| \mu |(\Omega) \leq 1 + \sup_n |\mu_n|(\Omega) < \infty$$.
Finally, to see that $$\mu_n \to \mu$$ for your norm, note that for $$\varepsilon > 0$$ there is an $$N$$ (independent of our choice of $$A_i$$) such that for $$n,m \geq N$$, $$\sum_i \|\mu_n(A_i) - \mu_m(A_i)\| \leq |\mu_n - \mu_m|(\Omega) \leq \varepsilon$$ Sending $$m \to \infty$$ on the left hand side and taking the $$\sup$$ gives for $$n \geq N$$ $$|\mu_n - \mu|(\Omega) \leq \varepsilon$$ which shows the desired convergence.
• In $\sum_{i=1}^k \|\mu(A_i)\| \leq 1 + \sum_i \|\mu_n(A_i)\| \leq 1 + \sup_n |\mu_n|(\Omega) < \infty$, how did you obtain the last inequality? Clearly, there is a $N\in\mathbb N$ with $\sum_{i=1}^k|\mu_N(A_i)-\mu(A_i)|<1$. Then $\sum_{i=1}^k|\mu(A_i)|\le\sum_{i=1}^k|\mu_n(A_i)-\mu(A_i)|+\sum_{i=1}^k|\mu_N(A_i)|<1+k|\mu_N(\Omega)|$. But this tends to $\infty$ as $k\to\infty$. – 0xbadf00d Jul 18 at 6:29
• $\sum_i |\mu_N(A_i)| \leq |\mu_N|(\Omega)$ without the $k$ there. This is literally just the definition of $|\mu_N|$. – Rhys Steele Jul 18 at 9:42
• Sure I forgot the definition for a moment. If each $\mu_n$ is even countable additive, is $\mu$ countable as well? – 0xbadf00d Jul 18 at 11:06
• $\mu$ is certainly countably additive if you're willing to assume that $\mathcal{A}$ is a $\sigma$-algebra. The argument I know does use that $\mathcal{A}$ is closed under complements so if you're not willing to assume that I'd have to think. Either way, the argument is too long to reproduce in the comments here. – Rhys Steele Jul 18 at 11:50